Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{77 x^{3}}{1000} - 8$ using $LaTeX: \displaystyle x_0=5$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{77 x_{n}^{3}}{1000} + 8 + e^{- x_{n}}}{- \frac{231 x_{n}^{2}}{1000} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 5$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (5.0000000000) - \frac{- \frac{77 (5.0000000000)^{3}}{1000} + 8 + e^{- (5.0000000000)}}{- \frac{231 (5.0000000000)^{2}}{1000} - e^{- (5.0000000000)}} = 4.7201080250$ $LaTeX: x_{2} = (4.7201080250) - \frac{- \frac{77 (4.7201080250)^{3}}{1000} + 8 + e^{- (4.7201080250)}}{- \frac{231 (4.7201080250)^{2}}{1000} - e^{- (4.7201080250)}} = 4.7029410951$ $LaTeX: x_{3} = (4.7029410951) - \frac{- \frac{77 (4.7029410951)^{3}}{1000} + 8 + e^{- (4.7029410951)}}{- \frac{231 (4.7029410951)^{2}}{1000} - e^{- (4.7029410951)}} = 4.7028786484$ $LaTeX: x_{4} = (4.7028786484) - \frac{- \frac{77 (4.7028786484)^{3}}{1000} + 8 + e^{- (4.7028786484)}}{- \frac{231 (4.7028786484)^{2}}{1000} - e^{- (4.7028786484)}} = 4.7028786476$ $LaTeX: x_{5} = (4.7028786476) - \frac{- \frac{77 (4.7028786476)^{3}}{1000} + 8 + e^{- (4.7028786476)}}{- \frac{231 (4.7028786476)^{2}}{1000} - e^{- (4.7028786476)}} = 4.7028786476$